$\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow XYZ$. $\overline{AD} \perp \overline{BC}$,$D \in \overline{BC}$ and $\overline{XM} \perp \overline{YZ}$,$M \in \overline{YZ}$. Prove that $\frac{AD}{XM} = \frac{BC}{YZ}$.

(N/A) Given: $\Delta ABC \sim \Delta XYZ$ with correspondence $ABC \leftrightarrow XYZ$. $\overline{AD} \perp \overline{BC}$ and $\overline{XM} \perp \overline{YZ}$.
To prove: $\frac{AD}{XM} = \frac{BC}{YZ}$.
Proof:
$1$. Since $\Delta ABC \sim \Delta XYZ$,the ratios of their corresponding sides are equal and corresponding angles are congruent.
$\therefore \frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$ and $\angle B \cong \angle Y$ ... $(1)$
$2$. In $\Delta ABD$ and $\Delta XYM$:
$\angle B \cong \angle Y$ (from result $1$)
$\angle ADB \cong \angle XMY$ (both are $90^{\circ}$ as $\overline{AD} \perp \overline{BC}$ and $\overline{XM} \perp \overline{YZ}$)
$3$. By the $AA$ similarity criterion,$\Delta ABD \sim \Delta XYM$.
$4$. Since the triangles are similar,the ratios of their corresponding sides are equal:
$\therefore \frac{AD}{XM} = \frac{AB}{XY}$ ... $(2)$
$5$. From $(1)$ and $(2)$,since $\frac{AB}{XY} = \frac{BC}{YZ}$,we get:
$\frac{AD}{XM} = \frac{BC}{YZ}$.
Hence proved.